Đề thi HK1 môn Toán lớp 10 Trường THPT Trần Phú -...

A \(y = {x^2} - 4x - 3\)

B \(y =  - {x^2} + 4x\)

C \(y = {x^2} + 4x - 3\)

D \(y =  - {x^2} + 4x - 3\)

A \(\overrightarrow x  = \left( { - 12;13} \right)\)

B \(\overrightarrow x  = \left( {12;13} \right)\)            

C \(\overrightarrow x  = \left( { - 13; - 12} \right)\)

D \(\overrightarrow x  = \left( { - 13;12} \right)\)

A \(x \ge 0;\,\,x \ne 1\)

B \(x \ge 1\)

C \(x > 1\)

D \(x > 0;\,\,x \ne 1\)

A 1

B 0

C 3

D 2

A \(\overrightarrow {AB}  = \left( {1; - 2} \right)\)

B \(\overrightarrow {AB}  = \left( { - 3;8} \right)\)

C \(\overrightarrow {AB}  = \left( {3; - 8} \right)\)

D \(\overrightarrow {AB}  = \left( { - 2; - 15} \right)\)

A 3

B 2

C 1

D 0

A \(1 \le m \le 3\)

B \(m \ge 3\)

C \(m < 1\)

D \(m > 3\)

A \(x =  - 2\)

B \(y =  - 4\)

C \(y = 2\)

D \(x = 2\)

A Hàm số đồng biến trên R

B Hàm số đồng biến trên khoảng \(\left( { - 2; + \infty } \right)\)

C Hàm số đồng biến trên khoảng \(\left( {2; + \infty } \right)\)

D Hàm số nghịch biến trên khoảng \(\left( { - 2; + \infty } \right)\)

A 0

B 1

C 2

D 3

A \(G\left( {8;7} \right)\)

B \(G\left( {\frac{8}{3};\frac{7}{3}} \right)\)

C \(G\left( { - \frac{8}{3}; - \frac{7}{3}} \right)\)

D \(G\left( { - \frac{8}{3};\frac{7}{3}} \right)\)

A \(y = {x^3} - x + 1\)

B \(y = {x^4} - 2{x^2} + 1\)

C \(y = \left| {x + 1} \right| + \left| {x - 1} \right

D \(y = 2x - {x^3}\)

A \(\left\{ {\frac{2}{{m + 3}}} \right\}\)

B \(\left\{ {\frac{2}{{m - 3}}} \right\}\)

C \(\emptyset \)

D \(R\)

A \(\left\{ {2 + \sqrt 2 } \right\}\)

B \(\emptyset \)

C \(\left\{ {2 + \sqrt 2 ;2 - \sqrt 2 } \right\}\)

D \(\left\{ {2 - \sqrt 2 } \right\}\)

A \(m > \sqrt 5 \)

B \(m \le \sqrt 5 \)

C \(m \ge \sqrt 5 \)

D \(m < \sqrt 5 \)

A \(m \ne 0;\,\,m \ne 2\)

B \(m \ne 2\)

C \(\forall m \in R\)

D \(m \ne 0\).

A \(R\)

B \(\left( { - 2; + \infty } \right)\)

C \(R\backslash \left\{ 2 \right\}\)

D \(R\backslash \left\{ { - 2} \right\}\)

A 3

B 4

C 1

D 6

A Ba điểm A, B, D

B Ba điểm A, B, C

C Ba điểm B, C, D

D Ba điểm A, C, D

A \(x=2;x=-1\)

B \(x=2\)

C \(x=-1\)

D Vô nghiệm

A \(\begin{array}{l}
a)\,\,S = \left\{ {\left( {2;5} \right)} \right\}\\
b)\,\, - \frac{5}{2} \le m \le \frac{5}{2}
\end{array}\)

B \(\begin{array}{l}
a)\,\,S = \left\{ {\left( {2;5} \right);\left( {3;5} \right)} \right\}\\
b)\,\, - \frac{5}{2} \le m \le \frac{5}{2}
\end{array}\)

C \(\begin{array}{l}
a)\,\,S = \left\{ {\left( {2;5} \right);\left( { - 3; - 5} \right)} \right\}\\
b)\,\, - \frac{5}{2} < m < \frac{5}{2}
\end{array}\)

D \(\begin{array}{l}
a)\,\,S = \left\{ {\left( {2;5} \right);\left( { - 3; - 5} \right)} \right\}\\
b)\,\, - \frac{5}{2} \le m \le \frac{5}{2}
\end{array}\)

A \(\begin{array}{l}
a)\,\,\overrightarrow {AB} .\overrightarrow {AC} = 12\\
b)\,\,\widehat {BAC} = {45^0}
\end{array}\)

B \(\begin{array}{l}
a)\,\,\overrightarrow {AB} .\overrightarrow {AC} =- 12\\
b)\,\,\widehat {BAC} = {45^0}
\end{array}\)

C \(\begin{array}{l}
a)\,\,\overrightarrow {AB} .\overrightarrow {AC} = 12\\
b)\,\,\widehat {BAC} = {135^0}
\end{array}\)

D \(\begin{array}{l}
a)\,\,\overrightarrow {AB} .\overrightarrow {AC} = 12\\
b)\,\,\widehat {BAC} = {90^0}
\end{array}\)

A \(\begin{array}{l}
a)\,\,\overrightarrow {AI} = \frac{1}{2}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AC} \\
\,\,\,\,\,\,\overrightarrow {AK} = \frac{1}{3}\overrightarrow {AC} \\
\,\,\,\,\,\,\overrightarrow {KI} = \frac{1}{2}\overrightarrow {AB} + \frac{1}{6}\overrightarrow {AC} \\
b)\,\,M \equiv I
\end{array}\)

B \(\begin{array}{l}
a)\,\,\overrightarrow {AI} = \frac{1}{2}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AC} \\
\,\,\,\,\,\,\overrightarrow {AK} = \frac{1}{3}\overrightarrow {AC} \\
\,\,\,\,\,\,\overrightarrow {KI} = \frac{1}{2}\overrightarrow {AB} + \frac{1}{6}\overrightarrow {AC}
\end{array}\)

b) M là trung điểm của AI

C \(\begin{array}{l}
a)\,\,\overrightarrow {AI} = \frac{1}{2}\overrightarrow {AB} - \frac{1}{2}\overrightarrow {AC} \\
\,\,\,\,\,\,\overrightarrow {AK} = \frac{1}{3}\overrightarrow {AC} \\
\,\,\,\,\,\,\overrightarrow {KI} = \frac{1}{2}\overrightarrow {AB} - \frac{1}{6}\overrightarrow {AC} \\
b)\,\,M \equiv I
\end{array}\)

D \(\begin{array}{l}
a)\,\,\overrightarrow {AI} = \frac{1}{2}\overrightarrow {AB} - \frac{1}{2}\overrightarrow {AC} \\
\,\,\,\,\,\,\overrightarrow {AK} = \frac{1}{3}\overrightarrow {AC} \\
\,\,\,\,\,\,\overrightarrow {KI} = \frac{1}{2}\overrightarrow {AB} - \frac{1}{6}\overrightarrow {AC}
\end{array}\)

b) M là trung điểm của AI